 # CHEN 4460 – Process Synthesis, Simulation and Optimization

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CHEN 4460 – Process Synthesis, Simulation and Optimization
Dr. Mario Richard Eden Department of Chemical Engineering Auburn University Lecture No. 8 – Mathematical Optimization October 16, 2012 Contains Material Developed by Dr. Daniel R. Lewin, Technion, Israel

Lecture 10 – Objectives On completion of this part, you should:
Understand the different types of optimization problems and their formulation Be able to formulate and solve a variety of optimization problems in LINGO

Optimization Basics What is Optimization? Examples
The purpose of optimization is to maximize (or minimize) the value of a function (called objective function) subject to a number of restrictions (called constraints). Examples 1. Maximize reactor conversion Subject to reactor modeling equations kinetic equations limitations on T, P and x

Optimization Basics Examples (Continued) 2. Minimize cost of plant
Subject to mass & energy balance equations equipment modeling equations environmental, technical and logical constraints

Optimization Basics Examples (Continued)
3. Maximize your grade in this course Subject to extracurricular activities full-time-job requirements constant demand by other courses and/or your advisor/boss

Optimization Basics Formulation of Optimization Problems
min (or max) f(x1,x2,……,xN) subject to g1(x1,x2,……,xN)≤0 g2(x1,x2,……,xN)≤0 gm(x1,x2,……,xN)≤0 h1(x1,x2,……,xN)=0 h2(x1,x2,……,xN)=0 hE(x1,x2,……,xN)=0 Feasibility Any vector (or point) which satisfies all the constraints of the optimization program is called a feasible vector (or a feasible point) The set of all feasible points is called feasibility region or feasibility domain Any optimal solution must lie within the feasibility region! Inequality Constraints Equality Constraints

Optimization Basics Classification of Optimization Problems
Linear Programs (LP’s) A mathematical program is linear if f(x1,x2,……,xN) and gi(x1,x2,……,xN)≤0 are linear in each of their arguments: f(x1,x2,……,xN) = c1x1 + c2x2 + …. cNxN gi(x1,x2,……,xN) = ai1x1 + ai2x2 + …. aiNxN where ci and aij are known constants. Linear Programs (LP’s) can be solved to yield a global optimum. Solver routines can guarantee a truly optimal solution.

Optimization Basics Classification of Optimization Problems
Non-Linear Programs (NLP’s) A mathematical program is non-linear if any of the arguments are non-linear. For example: min 3x + 6y2 s.t. 5x + xy ≥ 0 Integer Programming Optimization programs in which ALL the variables must assume integer values. The most commonly used integer variables are the zero/one binary integer variables. Integer variables are often used as decision variables, e.g. to choose between two reactor types. Non-Linear Programs (NLP’s) can be solved to yield a local optimum. Solver routines can not always guarantee a globally optimal solution.

Optimization Basics Classification of Optimization Problems
Mixed Integer Linear Programs (MILP’s) Linear programs in which SOME of the variables are real and other variables are integers Can be solved as individual LP’s by fixing the integer variables, thus a global optimum can be identified. Mixed Integer Non-Linear Programs (MINLP’s) Non-linear programs in which SOME of the variables are real and other variables are integers Can be solved as individual NLP’s by fixing the integer variables, but depending on the nature of the NLP’s it may not be possible to find a global optimum.

Optimization Basics Formulation of Optimization Problems Step 1 Step 2
Determine the quantity to be optimized and express it as a mathematical function (this is your objective function) Doing so also serves to define variables to be optimized (input variables or optimization variables) Step 2 Identify all stipulated requirements, restrictions, and limitations, and express them mathematically. These requirements constitute the constraints Step 3 Express any hidden conditions. Such conditions are not stipulated explicitly in the problem, but are apparent from the physical situation, e.g. non-negativity constraints

Optimization Example Hydrogen Sulfide Scrubbing
Two variable grades of MEA. First grade consists of 80 weight% MEA and 20% weight water. Its cost is 80 cent/kg. Second grade consists of 68 weight% MEA and 32 weight% water. Its cost is 60 cent/kg. It is desired to mix the two grades so as to obtain an MEA solution that contains no more than 25 weight% water. What is the optimal mixing ratio of the two grades which will minimize the cost of MEA solution (per kg)?

Variables (Basis 1 kg solution)
Optimization Example Hydrogen Sulfide Scrubbing (Cont’d) Objective function min z = 80x1 + 60x2 Constraints Water content limitation 0.20x x2 ≤ 0.25 Overall material balance x1 + x2 =1 Non-negativity x1 ≥ 0 x2 ≥ 0 Variables (Basis 1 kg solution) x1 Amount of grade 1 (kg) x2 Amount of grade 2 (kg) z Cost of 1 kg solution (cents)

Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region The set of points (x1, x2) satisfying all the constraints, including the non-negativity conditions. Constraint on water content x x2 ≤ 0.25

Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region Non-negativity constraints x1  0 , x2  0

Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region Mass balance constraint x1 + x2 = 1

Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Feasibility region Any optimal solution must lie within the feasibility region!

Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
By plotting objective function curves for arbitrary values of z (here 70 and 75) we can evaluate the results: Optimal Point Intersection between x1 + x2 = 1 and 0.20x x2 = 0.25 In addition 70 < zmin < 75

Optimization Example Hydrogen Sulfide Scrubbing (Cont’d)
Solving the two equations simultaneously yields the optimum amounts of the two MEA solutions along with the minimum cost of the mixture

Optimization Software
LINGO Available on computers in Ross 306 To start entering a new optimization problem type: Model: Enter the objective function by typing: min = ……; or max = ……; Then enter the constraints. Each line must end by a semi-colon ; The final statement in the problem should be “end”

Optimization Software
Resolving MEA Example in LINGO LINGO Input Model: min = 80*x1 + 60*x2; 0.2*x *x2 < 0.25; x1 + x2 = 1; x1 > 0; x2 > 0; end LINGO Output Rows= Vars= No. integer vars= ( all are linear) Nonzeros= Constraint nonz= 6( are +- 1) Density=0.667 Smallest and largest elements in absolute value= No. < : 1 No. =: 1 No. > : 2, Obj=MIN, GUBs <= 2 Single cols= 0 Optimal solution found at step: Objective value: Variable Value Reduced Cost X E+00 X E+00 Row Slack or Surplus Dual Price E E E+00 E+00 Value of objective function: Value of variable x1: Value of variable x2:

More Optimization Examples
Lab Experiment Determine the kinetics of a certain reaction by mixing two species, A and B. The cost of raw materials A and B are 2 and 3 \$/kg, respectively. Let x1 and x2 be the weights of A and B (kg) to be employed in the experiment The operating cost of the experiment is given by: OC = 4(x1)2 + 5(x2)2 The total cost of raw materials for the experiment should be exactly \$6. Minimize the operating cost!

More Optimization Examples
Lab Experiment (Cont’d) LINGO Input Model: min = 4*x1^2 + 5*x2^2; 2*x1 + 3*x2 = 6; x1 > 0; x2 > 0; end LINGO Output Rows= Vars= No. integer vars= Nonlinear rows= 1 Nonlinear vars= 2 Nonlinear constraints= 0 Nonzeros= Constraint nonz= Density=0.583 Optimal solution found at step: Objective value: Variable Value Reduced Cost X E+00 X E+00 Row Slack or Surplus Dual Price E E-07 E+00 Value of objective function: Value of variable x1: Value of variable x2:

More Optimization Examples
Coal Conversion Plant What are the optimal production rates of gaseous and liquid fuels that maximize the net profit of the plant? x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18

More Optimization Examples
Coal Conversion Plant (Cont’d) Objective function max z = 3x1 + 5x2 Constraints Pretreatment capacity 3x1 + 2x2 ≤ 18 Gasification capacity x1 ≤ 4 Liquefaction capacity 2x2 ≤ 12 Non-negativity x1 ≥ 0 x2 ≥ 0

More Optimization Examples Maximum profit Z = 36 for x1 = 2 and x2 = 6
Coal Conversion Plant (Cont’d) Graphical solution Maximum profit Z = 36 for x1 = 2 and x2 = 6

More Optimization Examples
Coal Conversion Plant (Cont’d) LINGO Input Model: max = 3*x1 + 5*x2; 3*x1 + 2*x2 <= 18; x1 <= 4; 2*x2 <= 12; x1 > 0; x2 > 0; end LINGO Output Rows= Vars= No. integer vars= ( all are linear) Nonzeros= Constraint nonz= 6( are +- 1) Density=0.611 Smallest and largest elements in absolute value= No. < : 3 No. =: 0 No. > : 2, Obj=MAX, GUBs <= 2 Single cols= 0 Optimal solution found at step: Objective value: Variable Value Reduced Cost X E+00 X E+00 Row Slack or Surplus Dual Price E+00 E E E+00 E+00 Value of objective function: 36 Value of variable x1: 2 Value of variable x2: 6

More Optimization Examples
Methanol Delivery Supply methanol for three Methyl acetate plants located in towns A, B, and C Daily methanol requirements for each plant: MeAc Plant location Tons/day A B C Methanol production plants MeOH plant Capacity

More Optimization Examples
Methanol Delivery (Cont’d) Shipping cost (100 \$/ton) Schedule the methanol delivery system to minimize the transportation cost MeOH Plant MeAc Plant A MeAc Plant B MeAc Plant C 1 2 5 3 8 11 6 15 4 7 9

More Optimization Examples
Methanol Delivery (Cont’d) We define the transportation loads (tons/day) going from each MeOH plant to each MeAc plant as follows: Total transportation cost (Z) MeOH Plant MeAc Plant A MeAc Plant B MeAc Plant C 1 X1A X1B X1C 2 X2A X2B X2C 3 X3A X3B X3C 4 X4A X4B X4C Z = 2X1A + X1B + 5X1C + 3X2A + 0X2B + 8X2C + 11X3A + 6X3B + 15X3C + 7X4A + X4B + 9X4C

More Optimization Examples
Methanol Delivery (Cont’d) Objective function min Z = 2X1A + X1B + 5X1C X2A + 0X2B X2C + 11X3A X3B + 15X3C X4A + X4B + 9X4C Constraints Availability/supply X1A + X1B + X1C = 7 X2A + X2B + X2C = 5 X3A + X3B + X3C = 3 X4A + X4B + X4C = 2 Requirements/demand X1A + X2A + X3A + X4A = 6 X1B + X2B + X3B + X4B = 1 X1C + X2C + X3C + X4C = 10

More Optimization Examples
Methanol Delivery (Cont’d) Constraints Non-negativity X1A ≥ 0 X1B ≥ 0 X1C ≥ 0 X2A ≥ 0 X2B ≥ 0 X2C ≥ 0 X3A ≥ 0 X3B ≥ 0 X3C ≥ 0 X4A ≥ 0 X4B ≥ 0 X4C ≥ 0

Mixed Integer Programs
Use of 0-1 Binary Integer Variables Commonly used to represent binary choices Dichotomy modeling

Mixed Integer Programs
The Assignment Problem Assignment of n people to do m jobs Each job must be done by exactly one person Each person can at most do one job The cost of person j doing job i is Cij The problem is to assign the people to the jobs so as to minimize the total cost of completing all the jobs. We can assign integer variables to describe whether a certain person does a certain job or not

Mixed Integer Programs
The Assignment Problem (Cont’d) The event of person j doing job i is designated Xij The objective function can be written as: Since exactly one person will do job i, and each person at most can do one job, we get:

Mixed Integer Programs
Plant Layout – An Assignment Problem Four new reactors R1, R2, R3 and R4 are to be installed in a chemical plant Four vacant spaces 1, 2, 3 and 4 are available Cost of assigning reactor i to space j (in \$1000) is Assign reactors to spaces to minimize the total cost Reactor Space 1 Space 2 Space 3 Space 4 R1 15 11 13 R2 12 17 R3 14 10 R4 16

Mixed Integer Programs
Plant Layout (Cont’d) Let Xij denote the existence (or absence) of reactor i in space j, i.e. if Xij =1 then reactor i exists in space j Objective function min Z = X X X X X X X X X X X X34 + 17X X X X44

Mixed Integer Programs
Plant Layout (Cont’d) Constraints Each space must be assigned to one and only one reactor X11 + X12 + X13 + X14 = 1 X21 + X22 + X23 + X24 = 1 X31 + X32 + X33 + X34 = 1 X41 + X42 + X43 + X44 = 1 Each reactor must be assigned to one and only one space X11 + X21 + X31 + X41 = 1 X12 + X22 + X32 + X42 = 1 X13 + X23 + X33 + X43 = 1 X14 + X24 + X34 + X44 = 1

Mixed Integer Programs
Plant Layout (Cont’d) Solve using LINGO Optimal assignment policy Reactor R1 in space 2 Reactor R2 in space 1 Reactor R3 in space 4 Reactor R4 in space 3 Minimum cost Cost = = \$49,000

Lecture 10 – Summary On completion of this part, you should:
Understand the different types of optimization problems and their formulation Be able to formulate and solve a variety of optimization problems in LINGO

Other Business Homework Next Lecture – October 23
SSLW: 24.1 plus problems posted on class webpage Due Tuesday October 23 LINGO software is available on class webpage as zip-file Next Lecture – October 23 Heat and Power Integration (SSLW p ) Review of Midterm Exam Thursday October 18 during lab sessions You will get your tests back to look at during solution review

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